Computing Limits at Infinity Part I

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Hckyplayer8

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Compute the lim as x --> infinity of [5x-3] / [9x+4]

lim as x --> infinity [(5x-3) / x ] / [(9x+4) / x] = lim as x --> infinity (5 - 3/x) / (9 + 4/x)

With x in the denominator, as x moving towards positive infinity ensures the fraction getting extremely small, thus the limit of both 3/x and 4/x is zero.

That leaves me with the answer 5/9.

Does that seem reasonable?
 
Yes, we can write:

[MATH]\frac{5x-3}{9x+4}=\frac{5-\dfrac{3}{x}}{9+\dfrac{4}{x}}[/MATH]
Both terms with \(x\) in the denominator will approach zero as \(x\) grows without bound, and so the limit is in fact 5/9.

edit: I see now that's what you wrote...my eyes tend to glaze over when presented with long strings of plain text math. :|
 
MarkFL said:
Yes, we can write:

[MATH]\frac{5x-3}{9x+4}=\frac{5-\dfrac{3}{x}}{9+\dfrac{4}{x}}[/MATH]
Both terms with \(x\) in the denominator will approach zero as \(x\) grows without bound, and so the limit is in fact 5/9.

edit: I see now that's what you wrote...my eyes tend to glaze over when presented with long strings of plain text math. :|
Click to expand...

Thank you.

Yeah...sometimes I feel guilty typing it out that way.

I'll be hanging out here for awhile since my degree pathway is math intensive. So I'll have to check out the math coding how to once this class is over. Sorry about the long plain text.

In regards to these type of problem, is it doing to just come down to the largest value in the numerator and denominator?

For example with the above. 5x and 9x were always going to be larger than the constants 3 and 4. So as x gets larger, those constants sort of become irrelevant and I could have just skipped to 5x/9x. Cancel out the variable and voila.

Or is that just a coincidence with this specific problem?
 
Hckyplayer8 said:
Thank you.

Yeah...sometimes I feel guilty typing it out that way.

I'll be hanging out here for awhile since my degree pathway is math intensive. So I'll have to check out the math coding how to once this class is over. Sorry about the long plain text.

In regards to these type of problem, is it doing to just come down to the largest value in the numerator and denominator?

For example with the above. 5x and 9x were always going to be larger than the constants 3 and 4. So as x gets larger, those constants sort of become irrelevant and I could have just skipped to 5x/9x. Cancel out the variable and voila.

Or is that just a coincidence with this specific problem?
Click to expand...

No worries about the plain text...that's just my old eyes barking...:D

Your observation is a good one, and is in some ways prescient regarding a method you will learn down the road called L'Hôpital's Rule. A consequence of this rule is that when you have a rational expression in which the degree of the polynomial in the numerator is equal to the degree of the polynomial in the denominator, then the limit at infinity will be equal to the ratio of the coefficients of the leading terms.
 
You can also think : What if I put in a large value for x, say 1 000 000, which makes the expression (5 000 000 -3 )/(9 000 000 + 4) which is not much different to 5 000 000/9 000 000 or 5/9.
 
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